Let d≥1 be an integer and let →A:={Ai}di=1 be real numbers. We consider a following integral:
I(d)(→A):=∞∫0e−u2[d∏i=1erf(Aiu)]du
By expanding the error functions in Taylor series and then integrating term by term we found the answer for d=1 and d=2. We have:
√πI(d)(→A)={arctan(A1)if d=1arctan(A1A2√1+A21+A22)if d=2
Now the question is how do we derive the result for arbitrary values of d?
Friday, 26 December 2014
integration - An integral involving error functions and a Gaussian
Subscribe to:
Post Comments (Atom)
real analysis - How to find limhrightarrow0fracsin(ha)h
How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
Ok, according to some notes I have, the following is true for a random variable X that can only take on positive values, i.e P(X \int_0^...
-
Self-studying some properties of the exponential-function I came to the question of ways to assign a value to the divergent sum $$s=\sum_{k=...
-
The question said: Use the Euclidean Algorithm to find gcd (1207,569) and write (1207,569) as an integer linear combination of 1207 ...
No comments:
Post a Comment